In Delta ABC,mangle B = 90^{circ} and overlin | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\Delta ABC$ is a right-angled tri

Vedclass · Accepted Answer

$20$

Vedclass · Answer

(C) In a right-angled triangle,the length of the median to the hypotenuse is half the length of the hypotenuse.Here,$\Delta ABC$ is a right-angled triangle with $m\angle B = 90^{\circ}$.Since $\overline{BM}$ is the median to the hypotenuse $\overline{AC}$,we have $BM = \frac{1}{2} AC$.Given $BM = 12.5$,so $12.5 = \frac{1}{2} AC$,which implies $AC = 25$.Now,using the Pythagorean theorem in $\Delta ABC$:$AB^2 + BC^2 = AC^2$$15^2 + BC^2 = 25^2$$225 + BC^2 = 625$$BC^2 = 625 - 225 = 400$$BC = \sqrt{400} = 20$.

In $\Delta ABC$,$m\angle B = 90^{\circ}$ and $\overline{BM}$ is a median. If $AB = 15$ and $BM = 12.5$,find $BC$.

Similar Questions

In $\Delta ABC,$ the bisector of $\angle A$ intersects $\overline{BC}$ at $D.$ If $AB = 6,$ $BD = 4$ and $DC = 3,$ find $AC$.

In $\Delta ABC$,$A-M-B$,$A-N-C$ and $\overline{MN} \parallel \overline{BC}$. If $\frac{AM}{MB} = \frac{2}{3}$ and $AC = 25$,find $NC$.

In $\Delta ABC$,$A-P-B$,$A-Q-C$ and $\overline{PQ} \parallel \overline{BC}$. If $\frac{AP}{PB} = \frac{3}{4}$ and $AC = 17.5$,find $AQ$.

In $\Delta XYZ$,$X-S-Y$,$X-T-Z$ and $\overline{ST} \parallel \overline{YZ}$. If $XS = 4$,$XT = 8$,$SY = x-4$ and $TZ = 3x-19$,find the value of $x$.

In $\Delta ABC$,$m \angle B = 90^{\circ}$ and $\overline{BM}$ is an altitude to the hypotenuse $\overline{AC}$. Then,$\angle BAM \cong \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module